A point charge `q` is placed at a distance `a//2` directly above the centre of a square of side `a`. The electric flux through the square is

Options

A. `(Qd^(2))/(6epsilon_(0))`
B. `(Qd)/(6epsilon_(0))`
C. `(Q)/(6epsilon_(0))`
D. `(6Q)/(epsilon_(0))`

Answer

C

Explanation

The given square of side d may be considered as one of the faces of a cube with edge d. Then given charge +Q will be considered to be placed at the centre of this cube. Then according to Gauss.s theorem, the magnitude of the electric flux through the faces (six) of the cube is given by `phi_(E) = Q//epsilon_(0)` Hence, electric flux through one face of the cube (or through the given square) will be `phi._(E) = (1)/(6) phi_(E) = (Q)/(6 epsilon_(0)`